COT 5405: Fall 2006 Lecture 26

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COT 5405: Fall 2006
Lecture 26
Maximum Flow Problem
Flow Networks
Definition: A flow network G=(V, E) is a directed graph where each edge (u, v) E has a
non-negative capacity c(u, v) 0.
o If (u, v) E, then we take c(u, v) = 0.
o Two special vertices, s, t V, s t, are denoted as the sink and source
respectively.
o For every v V, there is a path from s to t through v.
o So, |E| |V| - 1.
Definition: A flow is f: V V that satisfies the following constraints:
o Capacity constraint: u, v V , f(u v) c(u, v).
o Skew symmetry: u, v V , f(u, v) = -f(v, u).
o Flow conservation: u V – {s, t}, v V f(u, v) = 0.
Definition: The value of a flow f is defined as |f| = v V f(s, v).
Maximum flow problem: Given a flow network, find a flow of maximum value.
o Multiple sources/multiple sinks maximum flow problem can be solved by creating
a supersink (and a supersource) with edges to (and from) the sinks (and sources)
having infinite capacities.
Implicit summation notation: f(X, Y) = x X y Y f(x, y), when X, Y V.
Lemma 26.1
1. X V, f(X, X) = 0.
2. X, Y V, f(X, Y) = -f(Y, X).
3. X, Y, Z V with XY = , f(XY, Z) = f(X, Z) + f(Y, Z) and
f(Z, XY) = f(Z, X) + f(Z, Y).
Proof for #1 (the rest are review questions):
Label the vertices in X as x1, x2, …, xk.
f(X, X) = x X y X f(x, y) = ki=1 ji f(xi, xj) = ki=1 j>i f(xi, xj) + ki=1 j<i f(xi, xj)
= ki=1 j>i f(xi, xj) - ki=1 j<i f(xj, xi) = ki=1 j>i f(xi, xj) - kj=1 i>j f(xj, xi)
= ki=1 j>i f(xi, xj) -ki=1 j>i f(xi, xj) = ki=1 j>i ( f(xi, xj) - f(xi, xj) ) = 0.
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